Question

One mole of an ideal gas undergoes a change of state from (1.0 atm, 3.0 L, 200 K) to (4.0 atm, 5.0 L, 250 K) with a change in internal energy of 60 L-atm. What is the change in enthalpy of the process (in kJ)? (1 L-atm = 101 J)

• A. 6.66
• B. 7.77
• C. 7.07
• D. 8.88

Hint:

The change in enthalpy is given by \(\Delta H = \Delta U + \Delta (PV)\). Remember to convert units to be consistent throughout the calculation.

Answer 1

The Correct Option is B

1. Given data:

• Initial state: \( P_1 = 1.0 \) atm, \( V_1 = 3.0 \) L, \( T_1 = 200 \) K
• Final state: \( P_2 = 4.0 \) atm, \( V_2 = 5.0 \) L, \( T_2 = 250 \) K
• Change in internal energy (\( \Delta U \)) = 60 L-atm
• 1 L-atm = 101 J

2. Calculate the change in enthalpy (\( \Delta H \)):

\[ \Delta H = \Delta U + \Delta (PV) \] \[ \Delta H = \Delta U + (P_2V_2 - P_1V_1) \] \[ \Delta H = 60 \, \text{L-atm} + (4.0 \, \text{atm} \times 5.0 \, \text{L} - 1.0 \, \text{atm} \times 3.0 \, \text{L}) \] \[ \Delta H = 60 \, \text{L-atm} + (20 \, \text{L-atm} - 3 \, \text{L-atm}) \] \[ \Delta H = 60 \, \text{L-atm} + 17 \, \text{L-atm} \] \[ \Delta H = 77 \, \text{L-atm} \]

3. Convert L-atm to Joules (J):

\[ \Delta H = 77 \, \text{L-atm} \times 101 \, \text{J/L-atm} \] \[ \Delta H = 7777 \, \text{J} \]

4. Convert Joules to Kilojoules (kJ):

\[ \Delta H = \frac{7777 \, \text{J}}{1000 \, \text{J/kJ}} \] \[ \Delta H = 7.777 \, \text{kJ} \]

Therefore, the change in enthalpy of the process is 7.77 kJ.

Final Answer: 7.77 kJ.